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在将指针传递给需要常量引用的函数时，我看到了一些奇怪的行为。显然，C++ 希望我在将指针传递给期望引用的函数之前取消引用指针。我正在寻找错误或警告，但除非满足某个对我没有任何意义的条件，否则我不会收到错误或警告。

指针和引用的类型是 C++ 17 标准变体（如下）。如果我的 C++17 变体在模板中包含布尔值（如下所示），那么 GCC8.3 可以很好地编译代码。但我有意想不到的运行时结果。

如果我从标准变体模板中删除布尔值，则代码不会按我预期的那样编译。为什么有区别？

#include <iostream>
#include <variant>
#include <stdint.h>

typedef std::variant <
        std::monostate,
        int8_t,
        int16_t,
        int32_t,
        int64_t,
        uint8_t,
        uint16_t,
        uint32_t,
        uint64_t,
        float,
        double
        ,bool //If bool is in the variant, the compiler compiles the code without error and understandably has UB. Why isn't the lack of de-reference bug in DoThing not caught when that is the case?
> StdVARIANT;

//if 'bool' is commented out of the variant the error (expected) is:
//invalid initialization of reference of type ‘const StdVARIANT&’

static size_t GetIndexOfVariant(const StdVARIANT& RefToVariant);
static size_t DoThing(StdVARIANT* PtrToVariant);

static size_t DoThing(StdVARIANT* PtrToVariant){
    return GetIndexOfVariant(PtrToVariant); //this is a bug-PtrToVariant should be de referenced!
}

static size_t GetIndexOfVariant(const StdVARIANT& RefToVariant){
    size_t Index = RefToVariant.index();
    return Index;
}

int main()
{
    StdVARIANT* myvar = new StdVARIANT();
    *myvar = (int32_t)1;
    std::cout<<"long: "<<std::get<int32_t>(*myvar)<<" "<<sizeof(std::get<int32_t>(*myvar))<<std::endl;
    *myvar = (double)2;
    std::cout<<"double: "<<std::get<double>(*myvar)<<" "<<sizeof(std::get<double>(*myvar))<<std::endl;
    std::cout<< myvar->index() << " " << DoThing(myvar)<<std::endl; //when 'bool' is in the std::variant, these two calls return different results (UB)
    delete myvar;
return 0;
}

编译并运行为上面的代码块：

~/gcc-test$ g++ -std=c++17 -Wall -Wextra -pedantic main.cpp
~/gcc-test$ ./a.out
long: 1 4
double: 2 8
10 11

从 StdVARIANT 中注释 'bool'，并且：

~/gcc-test$ g++ -std=c++17 -Wall -Wextra -pedantic main.cpp
main.cpp: In function ‘size_t DoThing(StdVARIANT*)’:
main.cpp:27:30: error: invalid initialization of reference of type ‘const StdVARIANT&’ {aka ‘const std::variant<std::monostate, signed char, short int, int, long int, unsigned char, short unsigned int, unsigned int, long unsigned int, float, double>&’} from expression of type ‘StdVARIANT*’ {aka ‘std::variant<std::monostate, signed char, short int, int, long int, unsigned char, short unsigned int, unsigned int, long unsigned int, float, double>*’}
     return GetIndexOfVariant(PtrToVariant); //this is a bug-PtrToVariant should be de referenced!
                              ^~~~~~~~~~~~
main.cpp:23:15: note: in passing argument 1 of ‘size_t GetIndexOfVariant(const StdVARIANT&)’
 static size_t GetIndexOfVariant(const StdVARIANT& RefToVariant);
               ^~~~~~~~~~~~~~~~~

为什么我不总是得到错误，而只是在 bool 被注释掉时？

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1 回答 1

3

要重现该问题，您可以按如下方式简化代码

#include <iostream>
#include <variant>
#include <stdint.h>

typedef std::variant <
        std::monostate,
        int8_t,
        int16_t,
        int32_t,
        int64_t,
        uint8_t,
        uint16_t,
        uint32_t,
        uint64_t,
        float,
        double
        ,bool //If bool is in the variant, the compiler compiles the code without error and understandably has UB. Why isn't the lack of de-reference bug in DoThing not caught when that is the case?
> StdVARIANT;

int main()
 {
   StdVARIANT * pVar = new StdVARIANT();
   StdVARIANT   var = pVar;
 }

关键是 aStdVARIANT *不能转换为除 之外StdVARIANT的任何替代方案。bool

所以如果StdVARIANT包含bool,

   StdVARIANT   var = pVar;

var被初始化为包含一个bool值（ a true，假设pVar不是nullptr）。

如果StdVARIANT不包含bool，则var初始化失败。

您的DoThing()函数内部也是如此：当您调用时

GetIndexOfVariant(PtrToVariant);

临时StdVARIANT创建（或不创建，如果bool不可用）并将常量引用传递给GetIndexOfVariant().

于 2021-08-13T21:26:51.997 回答